Problem 184 - Calculus Volume 1

Question

Find \(\frac{d y}{d x}\) for the given functions. $$ y=\cos x(1+\csc x) $$

Step-by-Step Solution

Answer

\(-\sin x - 1 - \cos x \csc x \cot x\)

Step 1: Identify the Product Rule

Given the function \( y = \cos x(1 + \csc x) \), notice that it is a product of two functions: \( u = \cos x \) and \( v = (1 + \csc x) \). To differentiate this, we will use the product rule, which states \( \frac{d}{dx}[uv] = u'v + uv' \).

Step 2: Differentiate \( u = \cos x \)

Calculate the derivative of \( u = \cos x \). The derivative of \( \cos x \) is \( u' = -\sin x \).

Step 3: Differentiate \( v = 1 + \csc x \)

Calculate the derivative of \( v = 1 + \csc x \). The constant 1 has a derivative of 0, and the derivative of \( \csc x \) is \( v' = -\csc x \cot x \). Hence, \( v' = -\csc x \cot x \).

Step 4: Apply the Product Rule

Substitute \( u \), \( u' \), \( v \), and \( v' \) into the product rule formula: \( \frac{d}{dx}[uv] = (-\sin x)(1 + \csc x) + (\cos x)(-\csc x \cot x) \).

Step 5: Simplify the Expression

Expand and simplify the expression from the product rule step: \( \frac{d y}{d x} = -\sin x(1) - \sin x(\csc x) - \cos x \csc x \cot x \). This becomes: \( \frac{d y}{d x} = -\sin x - (1) - \cos x \csc x \cot x \), as \( \sin x \csc x = 1 \). Therefore, the simplified result is: \( \frac{d y}{d x} = -\sin x - 1 - \cos x \csc x \cot x \).

Key Concepts

Product Rule

The Product Rule is a fundamental tool in calculus used to differentiate expressions where two functions are multiplied together. Imagine you have two functions, say \( u(x) \) and \( v(x) \), being multiplied in a complex function. The product rule helps you find the derivative easily. When you see a multiplication, the product rule states:

• First, differentiate \( u \), call it \( u' \), while keeping \( v \) constant.
• Then, differentiate \( v \), call it \( v' \), while keeping \( u \) constant.
• Finally, plug into the formula: \( u'v + uv' \).

This combination respects the multiplication structure. For example, given \( y = \cos x(1 + \csc x) \), identify \( u = \cos x \) and \( v = 1 + \csc x \). Using the product rule ensures you evaluate all the derivatives necessary to get a complete answer for \( \frac{dy}{dx} \). When applying the product rule, remember to work through each component carefully to ensure accuracy in results.

Trigonometric Derivatives

Trigonometric functions have specific rules for differentiation which are crucial when tackling calculus problems involving sin, cos, tan, and other similar functions. Let's consider some basic derivatives, such as:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \( -\sin x \).
• For \( \csc x \), the derivative is \( -\csc x \cot x \).

These rules are derived from the unit circle and the limit definitions of derivatives. Knowing these derivatives by heart is essential. For example, in our main problem, differentiating \( v = 1 + \csc x \) requires these principles. The constant 1 disappears upon differentiation (as its derivative is 0), leaving the focus on \( \csc x \). Notice how the derivative becomes \( -\csc x \cot x \), a result of the inherent properties of trigonometric identities. It's vital to connect these derivatives back to the initial function they stem from for correct application.

Chain Rule

The Chain Rule is a powerful method used to differentiate composite functions. These are functions within functions, like peeling back layers of an onion. If you have a function natured as \( y = f(g(x)) \), the chain rule states the derivative is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). Here's how you break it down:

• Identify the outer function \( f(u) \) first, then the inner function \( u = g(x) \).
• Differentiate the outer function with respect to the inner, then multiply by the derivative of the inner function.

In practical scenarios, even simple expressions can need this rule. Though our original exercise focused on the product rule, recognizing when chain rule applies is crucial as products often embed composite relations. Imagine if inside those trigonometric functions, another layer exists, the chain rule would help unravel it effortlessly. To conclude, always confirm whether the functions are simple products or if further chains exist for complete derivatives. This method adds great flexibility and depth to solving advanced calculus problems efficiently.